The two possible solutions for the given triangle are:
Case 1: $B = 51.92°$, $C = 107.08°$, $c = 25.93$
Case 2: $B = 128.08°$, $C = 30.92°$, $c = 13.45$
Both solutions exist and are rounded to two decimal places.
The Law of Sines states that for any triangle ABC, the following equation holds:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Using this equation, we can solve for the missing angles and side lengths in the given triangle.
Case 1:
To find angle B, we can rearrange the equation to get:
$\sin B = \frac{b \sin A}{a}$
Plugging in the given values:
$\sin B = \frac{22 \sin 21°}{9.5}$
$\sin B = 0.789$
Taking the inverse sine of both sides:
$B = \sin^{-1}(0.789)$
$B = 51.92°$
To find angle C, we can use the fact that the sum of the angles in a triangle is 180°:
$C = 180° - A - B$
$C = 180° - 21° - 51.92°$
$C = 107.08°$
Finally, to find side c, we can use the Law of Sines again:
$\frac{c}{\sin C} = \frac{a}{\sin A}$
Rearranging and plugging in the given values:
$c = \frac{a \sin C}{\sin A}$
$c = \frac{9.5 \sin 107.08°}{\sin 21°}$
$c = 25.93$
So the solution for Case 1 is:
$B = 51.92°$, $C = 107.08°$, $c = 25.93$
Case 2:
In this case, we need to consider the possibility of an obtuse angle B. To find this angle, we can use the fact that the sine of an obtuse angle is the same as the sine of its supplement:
$\sin B = \sin (180° - B)$
So we can find the supplement of the angle we found in Case 1:
$B = 180° - 51.92°$
$B = 128.08°$
Plugging this value back into the Law of Sines equation, we can find the other missing values:
$C = 180° - A - B$
$C = 180° - 21° - 128.08°$
$C = 30.92°$
$c = \frac{a \sin C}{\sin A}$
$c = \frac{9.5 \sin 30.92°}{\sin 21°}$
$c = 13.45$
So the solution for Case 2 is:
$B = 128.08°$, $C = 30.92°$, $c = 13.45$
Therefore, the two possible solutions for the given triangle are:
Case 1: $B = 51.92°$, $C = 107.08°$, $c = 25.93$
Case 2: $B = 128.08°$, $C = 30.92°$, $c = 13.45$
Both solutions exist and are rounded to two decimal places.
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In the above rectangle, what is the length of each of the longer sides of the perimeter is 36?
Answer:
Step-by-step explanation:
2(2x+2) + 2(x+4) = 36
4x+4 +2x+8 =36
6x+12= 36
6x = 36-12
6x =24
x = 4
longer side = 10
HELPPPPPP PLEASEEEEE HURRYYYYY
68% of the races he competed in had a finish time around 64.5 and 65.5 seconds.
How to interpret a standard deviation?The term "variance" (or "") refers to an assessment of the data's dispersion from the mean. A small variance implies that the data are grouped around the normal, and while a large standard deviation shows that the data are more dispersed.
[tex]\begin{aligned}& \mathrm{P}(\mu-\sigma < \mathrm{X} < \mu+\sigma) \approx 68 \% \\& \mathrm{P}(\mu-2 \sigma < \mathrm{X} < \mu+2 \sigma) \approx 95 \% \\& \mathrm{P}(\mu-3 \sigma < \mathrm{X} < \mu+3 \sigma) \approx 99.7 \%\end{aligned}[/tex]
When the standard deviation out from mean of the distribution of X is and the mean of the dispersion of X is (assuming X is normally distributed).
Kiran's 400-meter dash timings have an average of 65 secs and a confidence interval of 0.5 seconds, and they are regularly distributed.
Using the formula, we then obtain
[tex]\mathrm{P}(65-0.5 < \mathrm{X} < 65+0.5)=\mathrm{P}(64.5 < \mathrm{X} < 65.5) \approx 68 \%[/tex]
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Susan opens a savings account with $900. She earned $54 in 1.5 years. What is the annual interest rate?
The annual interest rate on Susan's savings account is 4%.
What is the annual interest rate on Susan account?Annual interest rate, also known as annual percentage rate is the yearly interest generated by a sum that's charged to borrowers or paid to investors We can use the simple interest formula to calculate the annual interest rate which is Simple Interest = (Principal x Rate x Time)
By plugging our values, we get.
$54 = ($900 x Rate x 1.5)
Simplifying the equation:
Rate = $54 / ($900 x 1.5)
Rate = 0.04
Rate = 4%.
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What is E(Y | X<=1/2) ?
expectation of Y given that X is less than or equal half
The conditional expectation of Y given that X is less than or equal to 1/2 is calculated by taking the weighted average of the possible values of Y, with the weights being the probabilities of X being less than or equal to 1/2 for each value of Y.
The formula for E(Y | X<=1/2) is:
E(Y | X<=1/2) = ∑y P(Y=y | X<=1/2) * y
To find P(Y=y | X<=1/2), we can use the formula:
P(Y=y | X<=1/2) = P(X<=1/2 | Y=y) * P(Y=y) / P(X<=1/2)
We can then plug in the values for each possible value of Y and calculate the conditional expectation.
For example, if Y can take on the values 0, 1, and 2, and the probabilities of X being less than or equal to 1/2 for each value of Y are 0.2, 0.5, and 0.3, respectively, and the probabilities of Y being 0, 1, and 2 are 0.4, 0.3, and 0.3, respectively, then:
E(Y | X<=1/2) = (0.2 * 0.4 / 0.5) * 0 + (0.5 * 0.3 / 0.5) * 1 + (0.3 * 0.3 / 0.5) * 2
E(Y | X<=1/2) = 0 + 0.3 + 0.36
E(Y | X<=1/2) = 0.66
Therefore, the conditional expectation of Y given that X is less than or equal to 1/2 is 0.66.
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A rectangle is 45 cm long and 15 cm wide. What's the ratio of the rectangle's length to its width? A 1:3. B 3:1. C 1:4. D 4:1
Answer:
Area of a triangle =L*W
A= 45*15
A=270cm²
Solve for the missing variables. Please show your work.
In the triangle ABC, the value of x, y and z is obtained as 21, 7 and 48 units respectively.
What are triangles?
Triangles are a particular sort of polygon in geometry that have three sides and three vertices. Three straight sides make up the two-dimensional figure shown here. An example of a 3-sided polygon is a triangle. The total of a triangle's three angles equals 180 degrees. One plane completely encloses the triangle.
A triangle ABC is given.
The measure of AB is given as 16 + z units.
The measure of AD is given as 16 units.
The measure of DB is given as z - 16 units.
The measure of BE is given as 21 units.
The measure of BC is given as x units.
The measure of AC is given as 14 units.
According to the midpoint theorem, the length of DE is -
DE = 1/2 (AC)
y = 1/2 (14)
y = 7 units
Therefore, the value of y is obtained as 7 units.
Now according to indirect measurement -
AB / AC = BD / DE
Substitute the values in the equation -
16 + z / 14 = z - 16 / 7
7(16 + z) = 14(z - 16)
112 + 7z = 14z - 224
7z - 14z = -224 - 112
-7z = -336
z = 48
Therefore, the value of z is obtained as 48 units.
Now according to indirect measurement -
BC / AC = BE / DE
Substitute the values in the equation -
21 + x / 14 = 21 / 7
7(21 + x) = 14 × 21
147 + 7x = 294
7x = 294 - 147
7x = 147
x = 21
Therefore, the value of x is obtained as 21 units.
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Answer:
x = 21
y = 7
z = 32
Step-by-step explanation:
DE = 1/2 (AC)
y = 1/2 (14)
y = 7
AB / AC = BD / DE
Substitute values
16 + z / 14 = z - 16 / 7
7(16 + z) = 14(z - 16)
112 + 7z = 14z - 224
7z - 14z = -224 - 112
-7z = -336
z = 32
BC / AC = BE / DE
Substitute values
21 + x / 14 = 21 / 7
7(21 + x) = 14 × 21
147 + 7x = 294
7x = 294 - 147
7x = 147
x = 21
Ah Lee Arithmetic Operations on Functions Feb 20, 8:50:52 PM Given that f(x)=x^(2)-6x-40 and g(x)=x+4, find f(x)-g(x) and express the result as a polynomial in simplest form.
To find f(x)-g(x), we need to subtract the two given functions.
f(x) = x^(2)-6x-40
g(x) = x+4
f(x)-g(x) = (x^(2)-6x-40) - (x+4)
Next, we need to distribute the negative sign to the terms inside the parentheses:
f(x)-g(x) = x^(2)-6x-40 - x - 4
Then, we can combine like terms: f(x)-g(x) = x^(2)-7x-44
Therefore, the result of f(x)-g(x) is a polynomial in simplest form: x^(2)-7x-44.
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The rabbit population in a forest area grows at the rate of 8% monthly. If there are 200 rabbits in April, find how many rabbits(rounded to the nearest whole number) should be expected by next April. Use y=200(2.7)^0.08t.
The expected rabbit population by next April is 486.
The rabbit population in a forest area can be found using the formula y=200(2.7)^0.08t, where y is the expected population, t is the time in months, and 200 is the initial population. To find the expected population by next April, we need to plug in the value of t as 12 (since there are 12 months in a year) and solve for y.
y=200(2.7)^0.08t
y=200(2.7)^0.08(12)
y=200(2.7)^0.96
y=200(2.4315)
y=486.3
Since the question asks for the expected population rounded to the nearest whole number, we can round 486.3 to 486.
Therefore, the expected rabbit population by next April is 486.
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51% of a certain food is water. Out of 300 total grams, how many grams are not water?
Out of 300 grams, 147 grams of food is not water in the given food sample.
If 51% of a certain food is water, then 49% of it must be something other than water. To find out how many grams of the food are not water, we can first calculate how many grams of water there are,
300 grams x 51% = 153 grams of water
Then we can subtract this from the total weight to find the amount of food that is not water -
300 grams - 153 grams = 147 grams of food that is not water
Therefore, out of 300 grams of nutrition, 153 grams are water and 147 grams are not.
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Suppose ~u1, ~u2, . . . , ~uk∈Rnand let S = span( ~u1, ~u2, . . . , ~uk). Suppose ~x, ~y ∈S.
(a) Show that ~0 ∈S.
(b) Convert ~x ∈S and ~y ∈S into equations.
(c) Show that r~x ∈S for any scalar r.
(d) Show that ~x + ~y ∈S.
The we have shown that S is a subspace of Rn and that x + y ∈ S.
Suppose ~u1, ~u2, . . . , ~uk∈Rnand let S = span( ~u1, ~u2, . . . , ~uk). Suppose ~x, ~y ∈S.(d) Show that ~x + ~y ∈S.The solution is given as follows.The problem statement is given below.Suppose u1, u2, . . . , uk∈Rnand let S = span(u1, u2, . . . , uk). Suppose x, y ∈S. Show that x + y ∈S.To prove that x + y ∈ S, we must first prove that S is a subspace of Rn. We can then show that x + y is a linear combination of vectors in S, so it must be in S.Let's get started with the proof.Let S = span{u1, u2, . . . , uk} be a subspace of Rn. Then u1, u2, . . . , uk are linearly independent, which implies that they span S. Thus, any vector in S can be written as a linear combination of u1, u2, . . . , uk.Suppose x, y ∈ S. Then x = a1u1 + a2u2 + . . . + akuk, and y = b1u1 + b2u2 + . . . + bkuk, for some scalars a1, a2, . . . , ak and b1, b2, . . . , bk.Now we can show that x + y is a linear combination of vectors in S, so it must be in S.x + y = (a1 + b1)u1 + (a2 + b2)u2 + . . . + (ak + bk)ukSince a1, a2, . . . , ak and b1, b2, . . . , bk are scalars, (a1 + b1), (a2 + b2), . . . , (ak + bk) are also scalars. Thus, x + y is a linear combination of u1, u2, . . . , uk, and hence x + y ∈ S. Therefore, we have shown that S is a subspace of Rn and that x + y ∈ S.
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Module 06: Project Option 1
You guys got me on this right I need this done by 4:30 pm tomorrow please help me the assignment is below
An expression that looks like Sarah's expression but is not equivalent is:
10(5j + j + 3).
What can you say about expressions?A sentence qualifies as a mathematical expression if it comprises one or more mathematical operations, at least two numbers, or variables. Let's examine the writing style of expressions. A number is 6 times larger than the other number, x, by a factor of 2. This statement is represented mathematically by the expression x/2 + 6. Using mathematical expressions, complex puzzles can be solved.
now in the question,
We can write a look alike expression which is not equivalent is:
10(5j + j + 3).
Now translating the new expression into verbal:
Sarah gets paid $10 for every shark tooth she finds. After finding 3, she asked Jhon for help. When Jhon finds a tooth, by that time Sarah finds 5 of them.
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Big ideas 7.5 question
Answer:
66.5
Step-by-step explanation:
average the 2 together (57+76)/2
Show with calculations whether the 15 boxes of çremora will be enough to last for a year
As a result, the 15 cartons of creamer will last for 1,350 days, or nearly 3.7 expression years. This implies the consumer drinks one serving of creamer per day at a weight of 5 grammes per serving.
what is expression ?An expression in mathematics is a collection of representations, numbers, and conglomerates that mimic a statistical correlation or regularity. A real number, a mutable, or a mix of the two can be used as an expression. Mathematical operators include addition, subtraction, fast spread, division, and exponentiation. Expressions are often used in arithmetic, mathematics, and form. They are employed in the depiction of mathematical formulas, the solving of equations, and the simplification of mathematical relationships.
(Total amount of creamer in grammes) / number of days (Amount of creamer consumed per day in grams)
Let's start by calculating the entire amount of creamer in grammes:
(Number of cartons) x (Total quantity of creamer) (Amount of creamer per box)
15 boxes x 450 grammes each box = total quantity of creamer
Total creamer weight = 6,750 g
(Total amount of creamer in grammes) / number of days (Amount of creamer consumed per day in grams)
The number of days is 6,750 grammes divided by 5 grammes each day.
The number of days is 1,350.
As a result, the 15 cartons of creamer will last for 1,350 days, or nearly 3.7 years. This implies the consumer drinks one serving of creamer per day at a weight of 5 grammes per serving.
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-12 Correct answer=Brainly
Find the fourth proportion to 1 5/7, 2 3/14 and 3 3/5
The fourth proportion to 15/7,23/14, and 33/5 is 19.25.
To discover the fourth proportion, we require to set up a probability between the 3 presented figures and the fourth unknown variety, which we can name x.
23/1433/5 x
First, we need to transfigure all the mixed figures to unsuitable fragments
= (7/7 * 1)5/7 = 7/75/7 = 12/7
= (14/14 * 2)3/14 = 28/143/14 = 31/14
= (5/5 * 3)3/5 = 15/53/5 = 18/5
Now we're suitable to preference those values into the share
31/1418/5 x
To break for x, we're suitable to cross-multiply
12/7) * x = (31/14) *(18/5)
Simplifying
12x/ 7 = 279/14
Multiplying each aspects via7/12
x = (279/14) *(7/12)
x = 19.25
Thus, the fourth proportion to 15/7,23/14, and 33/5 is 19.25.
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5 out of the 10 employees at the water slides are temporary employees. What percentage of the employees at the water slides are temporary?
Write your answer using a percent sign (%).
Answer: 50%
Step-by-step explanation: There are 5 out of 10 employees at the water slide. 10 in this situation is going to be 100%
since there is only 5 out of 10 employees that are temporary
it would be 50%
Answer: 50%
Step-by-step explanation:
5. About What percent of students said they like Cookie Cake?
How many students answered this survey over their favorite Cake?
Types of Cake
Cookie Cake
Chocolate Cake
Vanilla Cake
Marble Cake
Number of
students
5
6
8
5. 20% of students said that they enjoy Cookie Cake.
6. 25 students answered a survey regarding their favorite cake.
What is the percentage?The percentage is defined as a ratio expressed as a fraction of 100.
The % formula is used to calculate a percentage of a whole in terms of 100.
Percentage = (Value/Total Value) × 100
Percentage of students said they like Cookie Cake:
5/(5 + 6 + 8 + 6) × 100%
5/25 × 100%
20%
The total number of students who answered this survey about their favorite Cake:
5 + 6 + 8 + 6
25
Thus, 25 students answered a survey regarding their favorite cake.
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6. Complete the frequency table and ogive for the number of heads flipped. (a) (Excel object) Determine the frequency of each of the indicated intervals. Make sure that these frequencies are entered i
To complete the frequency table and ogive for the number of heads flipped, we need to determine the frequency of each of the indicated intervals. We can do this by counting the number of times each interval appears in the data set and entering the frequencies into the table.
Here is how to do it step-by-step:
1. Start with the first interval, which is 0-4 heads. Count the number of times this interval appears in the data set. For example, if there are 3 occurrences of 0-4 heads, enter 3 in the frequency column for this interval.
2. Repeat this process for each of the remaining intervals, counting the number of occurrences and entering the frequencies into the table.
3. Once you have entered all the frequencies, you can create the ogive by plotting the cumulative frequencies on a graph. Start with the first interval and plot the cumulative frequency at the upper limit of the interval. For example, if the first interval is 0-4 heads and the frequency is 3, plot the point (4,3) on the graph.
4. Continue this process for each of the remaining intervals, adding the frequency of each interval to the cumulative frequency and plotting the point at the upper limit of the interval.
5. Once you have plotted all the points, connect them with a line to create the ogive.
Here is the completed frequency table and ogive for the number of heads flipped:
| Interval | Frequency |
|----------|-----------|
| 0-4 | 3 |
| 5-9 | 5 |
| 10-14 | 4 |
| 15-19 | 2 |
| 20-24 | 1 |
Ogive:
(4,3) (9,8) (14,12) (19,14) (24,15)
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Peter drove a total of 1000km and used 100 litres of petrol
Calculate the rate at which the petrol was used in kilometers per little
The rate at which the petrol was used in kilometers per liter is 10 km/L that means that for every litre of petrol used for driving, Peter's car could travel 10 kilometers.
Peter drove a thousand km and used 100 liters of petrol. To find the rate at which the petrol was utilized in kilometers consistent with liter, we divide the whole distance traveled via the quantity of petrol used.
The formulation for this is rate of petrol usage = overall distance traveled/quantity of petrol used.
Rate of petrol usage = total distance traveled/quantity of petrol used
Substituting the values we get,
= 1000 km / 100 L
= 10 km/L
This means that for every liter of petrol used, Peter was capable of travel 10 kilometers.
Thus, the rate at which the petrol was used in kilometers per liter is 10 km/L.
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27. \( \left\{\begin{array}{c}x+y+z=-1 \\ 2 x+3 y+2 z=3 \\ 2 x+y+2 z=-7 \\ x-3 y+2 z=10 \\ -x+3 y-z=-6 \\ -x+3 y+2 z=6 \\ 6 x-2 y+2 z=4 \\ 3 x-y+2 z=2 \\ -12 x+4 y-8 z=8\end{array}\right. \)
Using the Gaussian elimination method, the general solution is: [tex]\[ x = -31t + 6 \][/tex], [tex]\[ y = 5 \][/tex], [tex]\[ z = 31t \][/tex].
To solve this system of equations, we can use the Gaussian elimination method. This method involves creating a matrix with the coefficients of the equations and using row operations to reduce the matrix to a form that can be easily solved.
First, we create a matrix with the coefficients of the equations:
[tex]\[ \left( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 2 & 3 & 2 & 3 \\ 2 & 1 & 2 & -7 \\ 1 & -3 & 2 & 10 \\ -1 & 3 & -1 & -6 \\ -1 & 3 & 2 & 6 \\ 6 & -2 & 2 & 4 \\ 3 & -1 & 2 & 2 \\ -12 & 4 & -8 & 8 \end{array} \right) \][/tex]
Next, we use row operations to reduce the matrix to a form that can be easily solved:
[tex]\[ \left( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 5 \\ 0 & -1 & 0 & -5 \\ 0 & -4 & 1 & 11 \\ 0 & 4 & 0 & -5 \\ 0 & 4 & 1 & 7 \\ 0 & -8 & -4 & 10 \\ 0 & -4 & -1 & 5 \\ 0 & 8 & 4 & -4 \end{array} \right) \][/tex]
[tex]\[ \left( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 31 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 27 \\ 0 & 0 & -4 & 50 \\ 0 & 0 & -1 & 25 \\ 0 & 0 & 4 & -44 \end{array} \right) \][/tex]
[tex]\[ \left( \begin{array}{ccc|c} 1 & 0 & 1 & -6 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 31 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -4 \\ 0 & 0 & 0 & 174 \\ 0 & 0 & 0 & 56 \\ 0 & 0 & 0 & -168 \end{array} \right) \][/tex]
From this reduced matrix, we can see that there are infinitely many solutions to this system of equations. The general solution can be written as:
[tex]\[ x = -31t + 6 \][/tex]
[tex]\[ y = 5 \][/tex]
[tex]\[ z = 31t \][/tex]
Where t is any real number.
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A Box Contains 16 Silver Counters, 8 Brown Counters And 20 Pink Counters. What Is The Ratio Of Silver To Brown To Pink Counters In Its Simplest Form?
Answer:
4-2-5
Step-by-step explanation:
16, 8, and 20 can all be divided by 4
Leaving you with 4 2 and 5.
The ratio is 4 to 2 to 5
Which of the statements best describe the origin on the coordinate system? I. The x- and y-axes intersect at the origin. II. The origin is the distance from right to left. III. The point, (0 , 0), is the ordered pair at the origin. IV. The origin is the distance from top to bottom. A. I and III B. IV only C. I only D. II and IV
I WILL GIVE 50 POINTS
i will give brainy if anyone is willing to help me with this
Question 6
What is the interquartile range for the data set?
238, 240, 211, 233, 201, 221, 262, 201, 205, 224, 222, 253
Answer:
IQR - 31
You first need to find the median. After find the middle number in the numbers before and after your median. This will give you your first and third quartile. Add these together then divide by two. This will give your IQR.
(1×1,000)+(3×100)+(5×10)+(9×
10
1
)+(8×
100
1
)left parenthesis, 1, times, 1, comma, 000, right parenthesis, plus, left parenthesis, 3, times, 100, right parenthesis, plus, left parenthesis, 5, times, 10, right parenthesis, plus, left parenthesis, 9, times, start fraction, 1, divided by, 10, end fraction, right parenthesis, plus, left parenthesis, 8, times, start fraction, 1, divided by, 100, end fraction, right parenthesis
The value of the expression (1×1,000)+(3×100)+(5×10)+(9×1/10)+(8 x 1/100) is 1,350.98.
To find the value of the expression (1×1,000)+(3×100)+(5×10)+(9×1/10)+(8 x 1/100), we need to simplify and perform the arithmetic operations.
Multiplying each term in the expression, we get:
1 × 1,000 = 1,000
3 × 100 = 300
5 × 10 = 50
9 × 1/10 = 0.9
8 × 1/100 = 0.08
Adding up all the terms, we get:
1,000 + 300 + 50 + 0.9 + 0.08 = 1,350.98
In summary, to find the value of the expression, we multiply each term by its respective coefficient, then add up all the terms. In this case, the result is 1,350.98.
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Complete question is:
Find the value of the expression (1×1,000)+(3×100)+(5×10)+(9×1/10)+(8 x 1/100)?
Find i (the rate per period) and n (the number of periods)
for the following annuity.
Monthly deposits of $305 are made for 7 years into an annuity
that pays 8.5% compounded monthly.
Therefore, the rate per period (i) is 0.00708333 and the number of periods (n) is 84. The future value of the annuity is $34563.89.
To find i (the rate per period) and n (the number of periods) for the given annuity, we need to use the formula for the future value of an annuity:
FV = PMT × [(1 + i)^n - 1] / i
Where FV is the future value, PMT is the periodic payment, i is the rate per period, and n is the number of periods.
Given:
PMT = $305
i = 8.5% / 12 = 0.00708333 (since the interest is compounded monthly)
n = 7 × 12 = 84 (since there are 12 months in a year and the deposits are made for 7 years)
Plugging these values into the formula, we get:
FV = $305 × [(1 + 0.00708333)^84 - 1] / 0.00708333
FV = $305 × [1.80722 - 1] / 0.00708333
FV = $305 × 0.80722 / 0.00708333
FV = $34563.89
Therefore, the rate per period (i) is 0.00708333 and the number of periods (n) is 84. The future value of the annuity is $34563.89.
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Can someone help me out with the amount deprecating each year? I can’t figure it out.
9th grade algebra
Answer:
Year 1 Depreciation:$2,000
Depreciation Percentage:20.00%
Total Depreciation:$10,000
Final Year Depreciation:$2,000
Step-by-step explanation:
P.S by teacher told me the answer
Problem 7(a). Find the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \) 7 (b). Find the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra
The solution is \boxed{A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right]} and \boxed{A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right]}.
(a) To find the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \), we need to use the formula:
\( A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right] \)
where \( a=-3, b=0, c=0, d=5 \).
Plugging in the values, we get:
\( A^{-1} = \frac{1}{(-3)(5)-(0)(0)} \left[\begin{array}{rr}5 & 0 \\ 0 & -3\end{array}\right] \)
\( A^{-1} = \frac{1}{-15} \left[\begin{array}{rr}5 & 0 \\ 0 & -3\end{array}\right] \)
\( A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right] \)
So, the inverse of \( A=\left[\begin{array}{rr}-3 & 0 \\ 0 & 5\end{array}\right] \) is \( A^{-1} = \left[\begin{array}{rr}-\frac{1}{3} & 0 \\ 0 & -\frac{1}{5}\end{array}\right] \).
(b) To find the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra \), we need to use the formula:
\( A^{-1} = \frac{1}{\det(A)} \left[\begin{array}{rrr}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]^{T} \)
where \( a_{11}=4, a_{12}=0, a_{13}=0, a_{21}=0, a_{22}=-2, a_{23}=0, a_{31}=0, a_{32}=0, a_{33}=3 \) and \( \det(A) = (4)(-2)(3) - (0)(0)(0) - (0)(0)(0) = -24 \).
Plugging in the values, we get:
\( A^{-1} = \frac{1}{-24} \left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{array}\right]^{T} \)
\( A^{-1} = \frac{1}{-24} \left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{array}\right] \)
\( A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right] \)
So, the inverse of \( A=\left[\begin{array}{rrr}4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3\end{arra \) is \( A^{-1} = \left[\begin{array}{rrr}-\frac{1}{6} & 0 & 0 \\ 0 & \frac{1}{12} & 0 \\ 0 & 0 & -\frac{1}{8}\end{array}\right] \).
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Mrs. Brown has 11 more boys than girls in her class and has a total of 28 students. Which of the following systems of equations could be used to solve this problem?
A. B + 11 = G and B + G = 28
B. B = G – 11 and B + G = 28
C. G = B + 11 and B + G = 28
D. B = G + 11 and B + G = 28
B
B=G-11 and B+G=28 could be used to solve the problem
In the first race in the video, the student sprints at a speed of 10 yards per second. The math teacher only runs 3.75 yards per second. In the second race, the teacher is given a 10-yard head start, but their speeds remain the same. In the third race, the teacher is again given a 10-yard head start, but her speed is doubled. The student, however, continues to run at 10 yards per second. The length of each race is 100 yards. In each case, who will cross the finish line first? Will the student catch the math teacher, and if so when?
After considering the speeds of the student and the teacher and the distance they ran in each case, we found that the students wins in all the races.
What is meant by the speed of a body?Speed is defined as the ratio of distance travelled to the amount of time it took. As speed simply has a direction and no magnitude, it is a scalar quantity.
An object is considered to be moving at a uniform speed when it travels the same distance in the same amount of time.
When an object travels a different distance at regular intervals, it is said to have variable speed.
Average speed is the constant speed determined by the ratio of the total distance travelled by an object to the total amount of time it took to travel that distance.
Given,
The length of the race = 100 yards
1st race
The speed of student = 10 yards/s
The speed of teacher = 3.75 yards / s
Time taken by student = distance / speed = 100/10 = 10s
Time taken by teacher = 100 / 3.75 = 26.67 s
So the student finishes the race because he/she took less time.
2nd race
Distance run by teacher = 100 - 10 = 90
Speed of teacher = 3.75 yards / s
Time taken by teacher = 90/3.75 = 24s
The distance and speed of the student are the same.
Time taken by student = 10 s
Still, the student finishes the race first.
3rd race
Distance run by teacher = 100 - 10 = 90
Speed of teacher = 3.75 * 2 yards /s = 7.5 yards/s
Time taken by teacher = 90/ 7.5 = 12s
The distance and speed of the student are the same.
Time taken by student = 10 s
Still, the student finishes the race first.
Therefore by considering the speeds of the student and the teacher and the distance they ran in each case, we found that the students wins in all the races.
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a dog weighed 35 pounds last year. this year the dog weighs 55. round to nearest whole.
Answer:
Step-by-step explanation:
20